Problem: Solve the equation. $\dfrac{dy}{dx}=8x^3y-8xy$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=Ce^{2x^4-4x^2}$ (Choice B) B $y=e^{2x^4-4x^2}+C$ (Choice C) C $y=2x^4-4x^2+C$ (Choice D) D $y=C\left(2x^4-4x^2\right)$
We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=8x^3y-8xy \\\\ \dfrac{dy}{dx}&=y(8x^3-8x) \\\\ \dfrac{1}{y}\,dy&=(8x^3-8x)\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} \dfrac{1}{y}\,dy&=(8x^3-8x)\,dx \\\\ \int \dfrac{1}{y}\,dy&=\int (8x^3-8x)\,dx \\\\ \ln|y|&=2x^4-4x^2+C_1 \\\\ e^{\ln|y|}&=e^{2x^4-4x^2+C_1} \\\\ |y|&=e^{2x^4-4x^2}e^{C_1} \\\\ y&=Ce^{2x^4-4x^2} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=Ce^{2x^4-4x^2}$